﻿ Angle of Elevation and Depression Worksheet | Problems & Solutions

# Angle of Elevation and Depression Worksheet

Angle of Elevation and Depression Worksheet
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1.
A ladder with its foot on a horizontal flat surface rests against a wall. It makes an angle of 30° with the horizontal. The foot of the ladder is 41 ft from the base of the wall. Find the height of the point where the ladder touches the wall.
 a. 24 ft b. 23 ft c. 26 ft d. 29 ft

#### Solution:

Draw the figure for the given data.

The distance from the foot of the ladder to the foot of the wall = 41 ft
[Given.]

Let h be the height of the point A at which the ladder touches the wall.

From right triangle ABC, tan 30° = h41
Þ h = 41 tan 30° » 24 ft.
[Substitute tan 30° and simplify.]

The ladder touches the wall at a height of 24 ft.

2.
A man on the deck of a ship is 15 ft above sea level. He observes that the angle of elevation of the top of a cliff is 70° and the angle of depression of its base at sea level is 50°. Find the height of the cliff and its distance from the ship.
 a. 23 ft and 23 ft b. 238 ft and 41ft c. 50 ft and 13 ft. d. 33 ft and 20 ft.

#### Solution:

Draw the diagram.

Let PQ represent the height of the cliff.

Let RS represent the height of the ship.

Let y represent the distance between the cliff and the ship.

In right triangle PRS, tan 50° = 15y

y = 15tan 50° » 12.58
[Substitute tan 50° = 1.192 and simplify.]

In right triangle QTS, tan 70° = xTS = xy

x = y (tan 70°)

x = 12.58 (tan 70°)
[Substitute the value of y.]

x » 35
[Substitute the value of tan 70° and simplify.]

PQ = PT + QT = 15 + x = 15 + 35 = 50
[From the diagram.]

Therefore, the height of the cliff is 50 ft and its distance from the ship is 13 ft.

3.
The angle of elevation of the top of a tree is 30o from a point 28 ft away from the foot of the tree. Find the height of the tree rounded to the nearest feet.

 a. 23 ft b. 10 ft c. 16 ft d. 8 ft

#### Solution:

The measure of the angle of elevation from point A is 30.

In right triangle APQ, tan 30° = h28

h = 28 × 0.577
[Substitute the value of tan 30°.]
[Use calculator.]

h 16 ft
[Simplify.]

So, the height of the tree is 16 ft.

4.
From the top of a spire of height 50 ft, the angles of depression of two cars on a straight road at the same level as that of the base of the spire and on the same side of it are 25° and 40°. Calculate the the distance between the two cars.
 a. 47 ft b. 7.013 ft. c. 40.523 ft. d. 27.786 ft

#### Solution:

Draw the figure for the given data.

Let P and Q be the positions of the cars and x represent the distance between them.

Let y represent the distance of the car at P from the foot of the building.

In right triangle BAQ, tan 25° = 50y

y = 50tan 25°

y » 107
[Substitute the value of tan 25° and simplify.]

In right triangle BAP, tan 40° = 50 / AP

AP » 59.594
[Substitute the value of tan 40° and simplify.]

AP + PQ = 59.594 + PQ
[Substitute the value of AP.]

AQ = 59.594 + PQ
[From diagram AP + PQ = AQ.]

y = 59.594 + x

107 = 59.594 + x
[Substitute the value of y.]

x » 47

Therefore, the distance between the cars is 47 ft.

5.
From the top of a building of height $h$ meters in a street, the angles of elevation and depression of the top and the foot of another building on the opposite side of the street are $\theta$° and $\phi$° respectively. Find the height of the opposite building. [Given $h$ = 70, $\phi$ = 35, and $\theta$ = 50.]
 a. 214 meters b. 224 meters c. 229 meters d. 219 meters

#### Solution:

Draw the figure from the given data.

Let AB represent the first building and CE represent the opposite building.

CD = AB = 70 meters
[The opposite sides of a rectangle.]

In right triangle ABC, tan 35° = AB / BC

0.700 = 70BC
[Substitute the value of tan 35° and simplify.]

BC 100 meters

[Opposite sides of a rectangle.]

1.191 = DE100
[From step 7.]

DE 119.10 meter
[Substitute the value of tan 50° and simplify.]

Therefore, the height of the opposite building CE = CD + DE

100 + 119.10

219 meters.

Therefore, the height of the opposite building is 219 meters.

6.
From the top of a tower of height $h$ meters the angle of depression of a red car is $\phi$° and the angle of depression of a blue car which is on the opposite side of the tower is $\theta$°. If the distance between the foot of the tower and the red car is $y$ meters, then find the height of the tower and the distanc between the two cars.
Assume that the points of location of the cars and the foot of the tower are collinear. [Given $y$ = 115, $\theta$ = 55° and $\phi$ = 40°.]
 a. 96 m & 182 m b. 98 m & 183 m c. 96 m & 184 m d. 101 m & 182 m

#### Solution:

Draw the figure from the given data.

Let A represent the top of the tower.

Let BC represents the distance between two cars.

Let h represents the height of the tower and x represents the distance between the foot of the tower and the blue car.

In right triangle ABD, tan 40° = AD / BD = h115
Þ h = 115 tan 40°
Þ h » 96 meters.
[Substitute the value of tan 40° and simplify.]

In right triangle ADC, tan 55° = hx
Þ tan 55° = 96y
Þ y » 67 meters.
[Substitute the value of tan 55° and simplify.]

So, the distance between the two cars, BC = 115 + y = 115 + 67 = 182 meters.

7.
The angle of elevation of an unfinished tower from a point 120 m away from its base is 25°. How much higher the tower be raised so that its angle of elevation from the same point will be 40°?
 a. 50 m b. 47 m c. 48 m d. 45 m

#### Solution:

Draw the diagram.

The distance from the base of the unfinished tower to the point B = 120 meters.

Let the height of the tower to be raised to make an angle of 40° at point B be x meters.

Let the height of the unfinished tower, DC = h meters.

Consider right triangle BCD, tan 25° = DC / BC =h120

Þ h » 55 meters.
[Substitute the value of tan 25° and simplify.]

Consider right triangle ABC, tan 40° = AC / BC =x+h120

Þ x + h = 99.60
Þx + 55 = 99.60
[Substitute the value of tan 40° and simplify.]

Þ x » 45 meters.

So, the height of the tower to be raised to make the angle of elevation 40° at point B = 45 meters.

8.
The angle of elevation of the top of a cliff from the point Q on the ground is 30°. On moving a distance of 20 m towards the foot of the cliff the angle of elevation increases to $\phi$°. If the height of the cliff is 17.3 m, then find $\phi$.
 a. 45° b. 60° c. 120° d. 30°

#### Solution:

Draw the figure from the given data.

Height of the cliff, PA = 17.3 m.

Distance from point Q to point B = 20 m.

Let BA = x m.

In right triangle PQA, tan 30° = PA / QA =17.320+x

Þ 20 + x = 17.33
Þ x = 10 m.

Consider right triangle PBA, tan θ = PA / BA =17.310

Þ θ = 60°

9.
Tom and Sam are on the opposite sides of a tower of 160 meters height. They measure the angle of elevation of the top of the tower as 40° and 55° respectively. Find the distance between Tom and Sam.
 a. 308 m b. 313 m c. 303 m d. 306 m

#### Solution:

Height of the pole, AB = 160 m .

Let C & D be the position of Tom and Sam on either side of the tower.

tan 40° = 160BC
[tan R = AB / BC.]

BC » 160tan 40°
[Substitute the value of tan 40° and simplify.]

BC = 190.703 m

tan 55° = 160BD
[tan S = AB / BD.]

BD » 160tan 55° = 112.044 m
[Substitute the value of tan 55° and simplify.]

CD = CB + BD = 190.703 +112.044 » 303 m

So, the distance between Tom and Sam is 303 m.

10.
A man on the deck of a ship is 13 ft above water level. He observes that the angle of elevation of the top of a cliff is 40° and the angle of depression of the base is 20°. Find the distance of the cliff from the ship and the height of the cliff if the base of the cliff is at sea level.
 a. 36 ft and 45 ft. b. 36 ft and 43 ft. c. 18 ft and 43 ft. d. 18 ft and 21 ft

#### Solution:

Draw the diagram.

Let PQ represent the height of the cliff.

Let RS represent the height of the ship.

Let y represent the distance between the cliff and the ship.

In right triangle PRS, tan 20° = 13y

y = 13tan 20° » 35.714
[Substitute tan 20° and simplify.]

In right triangle QTS, tan 40° = xTS = xy

x = y (tan 40°)

x = 35.714 (tan 40°)
[Substitute the value of y.]

x » 30
[Substitute the value of tan 40° and simplify.]

PQ = PT + QT = 13 + x = 13 + 30 = 43
[From the diagram.]

Therefore, the distance of the cliff from the ship is 36 ft and the height of the cliff is 43 ft.